Understanding Feynman derivation of probability current density On page 21-4 of vol. III of The Feynman Lectures on Physics, he derives the formula for the probability current density. Starting with the Schrödinger equation, he presents the time derivative of the probability density, and then states,

"The potential terms and a lot of other stuff cancel out. And it turns out that what is left can indeed be written as a perfect divergence.”

However, this is true only if $\psi^*(A\cdot\nabla)\psi = -\psi(A\cdot\nabla)\psi^*$, which does not appear to be the case.  What am I missing?
 A: Feynman's book can be found at http://www.feynmanlectures.caltech.edu/III_21.html. 
You probably leave out some term. Just to give some hints:
\begin{equation}
\nabla \Bigg( \psi^* \Big(\frac{\hbar}{i} \nabla - q {\bf A} \Big) \psi \Bigg) = (\nabla \psi^*) \frac{\hbar}{i} (\nabla \psi) + \psi^* \frac{\hbar}{i} \nabla^2 \psi - \psi^* q \nabla({\bf A}\psi) - (\nabla\psi^*) q {\bf A} \psi
\end{equation}
\begin{equation}
\nabla \Bigg( \psi \Big(-\frac{\hbar}{i} \nabla - q {\bf A} \Big) \psi^* \Bigg) = -(\nabla \psi) \frac{\hbar}{i} (\nabla \psi^*) - \psi \frac{\hbar}{i} \nabla^2 \psi^* - \psi q \nabla({\bf A}\psi^*) - (\nabla\psi) q {\bf A} \psi^*
\end{equation}
Now use that $\nabla ({\bf A} \psi) = {\bf A} \nabla \psi$, because $\nabla {\bf A}=0$.
A: The question is how we simplify $\partial_t P=\frac{1}{i\hbar}(\psi^\ast\frac{1}{2m}(\frac{\hbar}{i}\nabla-qA)^2\psi+q\phi\psi^\ast\psi-\psi\frac{1}{2m}(-\frac{\hbar}{i}\nabla-qA)^2\psi^\ast-q\phi\psi\psi^\ast)$ to $\partial_t P=-\nabla(\frac{1}{2m}\psi^\ast(\frac{\hbar}{i}\nabla-qA)\psi+\frac{1}{2m}\psi(\frac{\hbar}{i}\nabla-qA)\psi^\ast)$, viz. Eqs. 21-10/11 here. The $\phi$ terms cancel trivially, so the difference between the two right-hand sides is $\frac{z+z^\ast}{2m}$ with $$z:=\frac{\psi^\ast(\frac{\hbar}{i}\nabla-qA)^2\psi}{i\hbar}+\nabla(\psi^\ast(\frac{\hbar}{i}\nabla-qA)\psi)=qA\psi^\ast(\nabla-\frac{iq}{\hbar}A)\psi+(\nabla\psi^\ast)(\frac{\hbar}{i}\nabla-qA)\psi.$$Alternatively, we can add any imaginary number we want to the definition of $z$ without changing $\frac{z+z^\ast}{2m}$, so e.g. we can put$$z=qA\psi^\ast\nabla\psi-qA(\nabla\psi^\ast)\psi,$$which by inspection is imaginary.
